Problem: $A=\left[\begin{array}{rr}-8 & -4 & 13 \\ 23 & 12 & -7 \\18 &10 &1\end{array}\right]$ $A_{1,3}=$
Answer: Background An $m\times n$ matrix has $m$ rows and $n$ columns. $A=\left[\begin{array}{rr}A_{1,1} & \cdots & A_{1,n} \\\\\vdots \ & \ddots & \vdots \\\\A_{m,1} &\cdots &A_{m,n}\end{array}\right]$ Therefore, the entry $A_{{c},{d}}$ is located on row ${c}$ and column ${d}$. Finding $A_{1,3}$ $A_{{1},{3}}$ is located on row ${1}$ of $A$ : $\left[\begin{array}{rr}\ {-8} & {-4} & {13} \\ 23 & 12 & -7 \\18 &10 &1\end{array}\right]$ $A_{{1},{3}}$ is also located on column ${3}$ of $A$. $\left[\begin{array}{rr}\ {-8} & {-4} &{\text{13}} \\ 23 & 12 & {-7} \\18 &10 & {1}\end{array}\right]$ Therefore, $A_{{1},{3}}={13}$. Summary $A_{1,3}=13$